Gegeben sei die Funktion: \(f(x) = {\left( {\cot \sqrt {2x} } \right)^2} = {\cot ^2}\left( {\sqrt {2x} } \right)\)
Bilde die Ableitungsfunktion f‘(x) gemäß den Regeln der Differentialrechnung.
\(f'\left( x \right) = - \dfrac{{\sqrt 2 }}{{\sqrt x }} \cdot \dfrac{{\cos \left( {\sqrt {2x} } \right)}}{{{{\sin }^3}\left( {\sqrt {2x} } \right)}} = - \dfrac{{\sqrt 2 }}{{\sqrt x }} \cdot \left[ {\cot \left( {\sqrt {2x} } \right) \cdot {{\csc }^2}\left( {\sqrt 2 } \right)} \right];\)
\(f'\left( x \right) = - \dfrac{{\sqrt 2 }}{{\sqrt {2x} }} \cdot \dfrac{{\cos \left( {\sqrt {2x} } \right)}}{{{{\sin }^3}\left( {\sqrt {2x} } \right)}} = - \dfrac{{\sqrt 2 }}{{\sqrt {2x} }} \cdot \left[ {\cot \left( {\sqrt {2x} } \right) \cdot {{\csc }^2}\left( {\sqrt 2 } \right)} \right];\)
\(f'\left( x \right) = - \dfrac{1}{{\sqrt x }} \cdot \dfrac{{\cos \left( {\sqrt {2x} } \right)}}{{{{\sin }^3}\left( {\sqrt {2x} } \right)}} = - \dfrac{1}{{\sqrt x }} \cdot \left[ {\cot \left( {\sqrt {2x} } \right) \cdot {{\csc }^2}\left( {\sqrt 2 } \right)} \right];\)
Ich errechne eine abweichende Lösung